if there is a jar containing 5 red marbles 6green and 4 blue what is the probability off not chossing a blue marble
5:16
It is not explicitly stated in the question, but it is assumed that you draw one marble from each bag. In this case, you have unrelated sequential probability, similar to tossing three coins. The answer is 0.53 or 0.125.
7/(4+7+5) = 7/16 = 43.75%
11 marbles total and 6 are blue so probability is 6/11
Number of different possible choices = 8 + 6 + 9 = 23Number of available successful choices (blue marbles) = 6Probability of success = 6/23 = 0.26087 = 26.087 %(rounded)
It can not be determined with the data provided.
The probability of choosing a blue marble is 5 in 15 or 1 in 3. The probability of then choosing a green marble is 5 in 14. (One is missing) Multiply the two probabilities and you get 5 in 42.(P = 0.1190... about 12%).
100%
it depends how many blue marbles there are and how many marbles total.
This is the same as the probability of choosing either a red of a blue marble. There are 5+4 out of 15 ways of doing this. The probability is therefore 9/15 = 3/5.
1/3 or 33%
cant answer this question without more information....probability requires a ratio.
5:16
We will assume that the bag contains 11 marbles (this is not a trick question!), so 4 are blue and 7 are not blue, so the probability of a color other than blue is 7/11.
The answer depends on how many blue and non-blue marbles there are, whether the choice is random and how many marbles are chosen. There is no information provided on any of these.
If one marble is chosen at random, the probability is 6/(4+6+5) = 6/15 = 2/5
The probability of choosing a green marble from this jar would be 6/15. You get this answer by adding up the sum of all the marbles.